package com.zs.letcode.top_interview_question_medium;

import java.util.LinkedList;
import java.util.Queue;

/**
 * 岛屿数量
 * 给你一个由'1'（陆地）和 '0'（水）组成的的二维网格，请你计算网格中岛屿的数量。
 * <p>
 * 岛屿总是被水包围，并且每座岛屿只能由水平方向和/或竖直方向上相邻的陆地连接形成。
 * <p>
 * 此外，你可以假设该网格的四条边均被水包围。
 * <p>
 * 示例 1：
 * <p>
 * 输入：grid = [
 * ["1","1","1","1","0"],
 * ["1","1","0","1","0"],
 * ["1","1","0","0","0"],
 * ["0","0","0","0","0"]
 * ]
 * 输出：1
 * 示例 2：
 * <p>
 * 输入：grid = [
 * ["1","1","0","0","0"],
 * ["1","1","0","0","0"],
 * ["0","0","1","0","0"],
 * ["0","0","0","1","1"]
 * ]
 * 输出：3
 *
 * <p>
 * 提示：
 * <p>
 * m == grid.length
 * n == grid[i].length
 * 1 <= m, n <= 300
 * grid[i][j] 的值为 '0' 或 '1'
 * 相关标签
 * 深度优先搜索
 * 广度优先搜索
 * 并查集
 * 数组
 * 矩阵
 * <p>
 * 作者：力扣 (LeetCode)
 * 链接：https://leetcode-cn.com/leetbook/read/top-interview-questions-medium/xvtsnm/
 * 来源：力扣（LeetCode）
 * 著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。
 *
 * @author madison
 * @description
 * @date 2021/10/11 15:35
 */
public class Chapter15 {
    public static void main(String[] args) {

    }

    private class Solution {
        /**
         * 方法一：深度优先搜索
         *
         * @param grid
         * @return
         */
        void dfs(char[][] grid, int r, int c) {
            int nr = grid.length;
            int nc = grid[0].length;

            if (r < 0 || c < 0 || r >= nr || c >= nc || grid[r][c] == '0') {
                return;
            }

            grid[r][c] = '0';
            dfs(grid, r - 1, c);
            dfs(grid, r + 1, c);
            dfs(grid, r, c - 1);
            dfs(grid, r, c + 1);
        }

        public int numIslands(char[][] grid) {
            if (grid == null || grid.length == 0) {
                return 0;
            }

            int nr = grid.length;
            int nc = grid[0].length;
            int num_islands = 0;
            for (int r = 0; r < nr; ++r) {
                for (int c = 0; c < nc; ++c) {
                    if (grid[r][c] == '1') {
                        ++num_islands;
                        dfs(grid, r, c);
                    }
                }
            }

            return num_islands;
        }

        /**
         * 方法二：广度优先搜索
         */
        public int numIslands1(char[][] grid) {
            if (grid == null || grid.length == 0) {
                return 0;
            }

            int nr = grid.length;
            int nc = grid[0].length;
            int num_islands = 0;

            for (int r = 0; r < nr; ++r) {
                for (int c = 0; c < nc; ++c) {
                    if (grid[r][c] == '1') {
                        ++num_islands;
                        grid[r][c] = '0';
                        Queue<Integer> neighbors = new LinkedList<>();
                        neighbors.add(r * nc + c);
                        while (!neighbors.isEmpty()) {
                            int id = neighbors.remove();
                            int row = id / nc;
                            int col = id % nc;
                            if (row - 1 >= 0 && grid[row - 1][col] == '1') {
                                neighbors.add((row - 1) * nc + col);
                                grid[row - 1][col] = '0';
                            }
                            if (row + 1 < nr && grid[row + 1][col] == '1') {
                                neighbors.add((row + 1) * nc + col);
                                grid[row + 1][col] = '0';
                            }
                            if (col - 1 >= 0 && grid[row][col - 1] == '1') {
                                neighbors.add(row * nc + col - 1);
                                grid[row][col - 1] = '0';
                            }
                            if (col + 1 < nc && grid[row][col + 1] == '1') {
                                neighbors.add(row * nc + col + 1);
                                grid[row][col + 1] = '0';
                            }
                        }
                    }
                }
            }

            return num_islands;
        }
    }
}
